Find minimum of f(x) = x^2

Know that f'(x) = 2*x = 0 so 

then 0/2 = 0

finally the answer is minimum of x^2 is 0.

Next homework help data.

so y'(x) = 2*x + 7 -2*3*x^-3.

set it equal to zero

0 = 2*x+7-6*x^-3

0*x^3=2*x^5+7*x^3-6

0=2*x^5+7*x^3-6.

solve for x. see Maple work